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Author: Subject: Ammonium nitrate detonation through oxypropane
dann2
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[*] posted on 25-6-2011 at 16:06
ATFQ




The original question has only been answered (probably speculation) my one person in the entire thread. (Thank you freedompyro). Dann2 said he haddent a clue.

We have had a discussion regarding:
Difficulty of removing valves from Propane tanks.
Cost of Proane tanks.
Sanity of peoply who may possible try out the idea.
Bunker building.
% effectiveness of entire operation.
Colour of necks of people (allegedly) trying out the idea.
Difficulty of pouring dry prills into an (approx.) one inch hole (ie.valve removed).
Substitution of tannerite (instead of AN).
Lack of exact figures/parameters from original Uncle.
Legality of refilling Propane tanks (with AN? !).
Phosphorus making with empty Propane tank.
Superman contacts.
Role (or lack thereof) of idea in professional military or civilian arsenals.
Size of Propane tanks.

SUCH EVASION!!!!!!!


Perhaps the original question simply cannot be answered unless by actually building the thing.
It would be dangerous of course unless a long line was used between AN filled (buried in site of pond* to be) tank and the Propane source.
The Propane to be pumped into the buried cylindar containing AN is gasous Propane, not liquid Propane.

It's not that incredabely hard to set up.
Remove valve from top of large (EMPTY) Propane cylindar (or any gas cylindar).
This is easy to do.
Fit one gas line to cylindar (drill hole and weld solder).
This is easy to do.
Fit ignition device (spark plug, hot wire, or whatever).
This is easy to do.
Fill cylindar with AN prills using the approx. one inch hole on top of cylindar where valve used to be.
This is easy to do.
Replace top valve.
This is easy to do.
Pump in (say pure Oxygen) using Oxygen cylindar to a pressure of (say 20PSI).
This is easy to do and things are still safe (IMHO).
Bury and get out the long gas line and get to a safe distance at end of line (bunker).
This is easy to do.
Pump in Propane (or other gas) untill proper stioc. amount for gas explosiong had flowed.
This is easy to do.
Let pond maker sit 24 hours (is this necesary?)
Ignite device

WILL THE AMMONIUM NITRATE DETONATE?
That was the original question.

The cylindar will rupture and you may only get a one duck pond contaminated with AN?

Doe's anyone here know the answer or can give an actual educated guess (without discussing all the above details\points as an evasion to the question).

Dann2 has not got a clue but would be inclined to guess that the AN would not detonate.

It's all a bit too Kewl for SM I will admit.

Dann2


*duck or otherwise
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[*] posted on 25-6-2011 at 16:40


Quote: Originally posted by dann2  


The original question has only been answered (probably speculation) my one person in the entire thread. (Thank you freedompyro). Dann2 said he haddent a clue.

We have had a discussion regarding:
Difficulty of removing valves from Propane tanks.
Cost of Proane tanks.
Sanity of peoply who may possible try out the idea.
Bunker building.
% effectiveness of entire operation.
Colour of necks of people (allegedly) trying out the idea.
Difficulty of pouring dry prills into an (approx.) one inch hole (ie.valve removed).
Substitution of tannerite (instead of AN).
Lack of exact figures/parameters from original Uncle.
Legality of refilling Propane tanks (with AN? !).
Phosphorus making with empty Propane tank.
Superman contacts.
Role (or lack thereof) of idea in professional military or civilian arsenals.
Size of Propane tanks.

SUCH EVASION!!!!!!!



You forgot:
1) Autoignition temperature of pressurised propane/oxygen mixes
2) Whether additives, dirt, AN or impurities would cause spontaneous ignition during filling
3) Oxyliquit in lieu of oxygen-propane.

Evasion is one thing ignoring the safety aspects is another. Might I add to your list there are just so many things that can go wrong with the propane/oxygen filling operation...
1) Pressurise the gases too fast - temperature rises until autoignition temperature is reached
2) Pressure lines are too short or insulated. No cooling effect of long hoses.
3) Compressor is too powerful - gases come in hot and don't have time to be compressed isothermally.
4) Impurities in the AN (presumably dirty fertiliser grade), tank, lubricants, seals, etc. which may spontaneously ignite at a lower temperature than an oxy-propane mix. E.g. diesel has an autoignition temperature of 210 C and butane 405 C.

I am not a puppetmaster. If someone wants to indulge in risky activities then it is their business. But don't say they weren't informed of what could go wrong.




I can sell the following:
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2) Alkex para-aramid Korean Kevlar analogue fabric (about 50% Du Pont's prices)
3) NdFeB magnets
4) High purity technical ceramics
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Bot0nist
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[*] posted on 25-6-2011 at 16:59


Couldn't this principle be replicated on a much smaller scale more safely, say in a home built device or a re-purposed CO<sub>2</sub> tank for a paint ball gun and using a sand pulverizing test to determine if AN detonation occurs?

[Edited on 26-6-2011 by Bot0nist]




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[*] posted on 25-6-2011 at 18:23


Hydrocarbons are quite finicky about having a close to stociometric ratio in order to burn. I seen to remember that it must be between 4 and 14% for natural gas to air.
2 to 1 air to propane, as quoted in the original post, is too rich. You would need to do
quite a bit of tweeking to get the right ratio.

A propane tank is good for a bout 100Psi. (Around 5 ATM). Maybe the ratio is less critical at higher pressures??

With the right ratio, the fuel air should detonate. I built a propane powered potato cannon a while back and it seemed to show detonation at atmospheric pressure.
(good cannons have a propane metering device, a small fan to mix the gas
and use a BBQ sparker with multiple spark gaps in series for faster ignition.

However it seems like the pressure produced by the burning gas will be much to low
to detonate the AN. About 1000Psi for the propane/air -vs- 20,000 Psi for a detonating HE.
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[*] posted on 25-6-2011 at 23:57


Quote: Originally posted by Bot0nist  
Couldn't this principle be replicated on a much smaller scale more safely, say in a home built device or a re-purposed CO<sub>2</sub> tank for a paint ball gun and using a sand pulverizing test to determine if AN detonation occurs?


Yes it can be replicated on a smaller scale, but that is why I said an oxypropane mix has no place in civilian or military arsenals...because even if you show it can detonate AN there are much safer and more convenient ways of doing so even for a home enthusiast without access to detonators.

The only way it'd be safe to use oxypropane explosions is by having a binary warhead system with mixing in situ, or a concentric tank arrangement where one tank is burst into another tank by a HE burster charge.




I can sell the following:
1) Various high purity non-ferrous metals - Ni, Co, Ta, Zr, Mo, Ti, Nb.
2) Alkex para-aramid Korean Kevlar analogue fabric (about 50% Du Pont's prices)
3) NdFeB magnets
4) High purity technical ceramics
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dann2
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[*] posted on 26-6-2011 at 06:06



The tank + contents heating up when gas is pumped in would not be a problem. After air or Oxygen is pumped in (at this stage the system is still safe enough to sit on IMO assuming cylindar is reasonable clean to start with) the tank can be let cool down. The system will heat up damm all anyways. The pressure in the tank will have increased to perhaps 20 or 40 PSI. Very little gas (O2) will have been pumped into system in order to get this increase in pressure as there is very little room in the cylindar (its filled with AN prills). A few liters of gas will have increased in pressure from atmospheric to 20/40 PSI. This will cause very very little heating. (few degreese)
If putting in O2 from an O2 cylinder, the O2 will have DECREASED in pressure (from perhaps a few thousand PSI to 20/40 PSI), therefor the pond making device will have COOLED. (rolls eyes here).


Start adding Propane gas (using long pipe). The Propane gas has to come out of another cylinder which is at approx. 100 PSI.
(177 psi of pressure to keep Propane liquid at 37.8 °C (from Wiki)) Once again the gas is going from higher pressure to lower pressure (say 40 psi in pond maker). The pond maker will COOL.(more roll eyes).


The exact ratio of O2 to Propane I am not addressing and would (IMO) would be the hardest part of the whole operation to get right and implement.

If doing the thing right you would put a vacuum onto the pond maker first for some hours so that flammable mixture would impregnate prills and there would be more room in cylindar.

Of course the whole scheme is a bit barmy. Most things in life are.
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[*] posted on 26-6-2011 at 08:47


Quote: Originally posted by gregxy  

A propane tank is good for a bout 100Psi. (Around 5 ATM).
...
However it seems like the pressure produced by the burning gas will be much to low
to detonate the AN. About 1000Psi for the propane/air -vs- 20,000 Psi for a detonating HE.


On the first count, no. Burst pressure for 20lb propane cylinders is on the close order of 1000psi. If they were only good for 100psi they'd explode when left out in the sun; propane vapor pressure exceeds 100psi somewhere around your body temperature IIRC.
On the second count, while I have no specific information regarding pressures inside the tank once ignited, it's clear that because the reacting gases will be at a much higher density than they would be at STP that your figures for what happens in a propane cannon at 1 bar are not going to be correct for an explosion inside a propane tank.
I'm somewhat curious where you can do this that the neighbors wouldn't complain. Frankly even if you had a large ranch I expect that the small seismic event might eventually have the Feds stopping by to see what was going on.
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[*] posted on 26-6-2011 at 12:43


Quote: Originally posted by bbartlog  
On the first count, no. Burst pressure for 20lb propane cylinders is on the close order of 1000psi. If they were only good for 100psi they'd explode when left out in the sun; propane vapor pressure exceeds 100psi somewhere around your body temperature IIRC.
On the second count, while I have no specific information regarding pressures inside the tank once ignited, it's clear that because the reacting gases will be at a much higher density than they would be at STP that your figures for what happens in a propane cannon at 1 bar are not going to be correct for an explosion inside a propane tank.
I checked the safety valve on my medium sized propane tank, it's 375 psi. This is the DOT-rated tank pressure. For ASME rated tanks, it's 240 psi. My 50 psi estimate above was way too low and made in haste. Burst pressures are rated nominally 3-4 times the relief valve. Tested ultimate failure is 1200 psi and above.

The burst pressure, though, is only part of the story. What's far more relevant is the maximum pressure achieved during the propane ignition burst. Taking a VOD of ammonium nitrate (AN) at a nominal 3000 m/s and the dimension of the charge at 30 cm, such an HE charge takes 0.1 ms to fully propagate. So if the tank can achieve, say, 4,000 psi for a whole millisecond before the pressure starts to fall, there's plenty of time for the HE charge to completely detonate. The pressure/temperature to initiate detonation is at most its propagation P/T, but I couldn't say how much less.

So suppose the tank is pressurized at 200 psi. Is it feasible to think that it could get to 4,000 psi, a 20x pressure multiplier? Estimate the flame temperature at 1200 K vs. nominal 300 K for ambient. So you can get 4x just on temperature. The other 5x would have to come from increases in the number of molecules. The gas mixture is oxygen-deficient, but at the temperatures of combustion, the AN is going to melt, and you're going to have reactive, liquid coated prills. AN is an oxidizer, so you can assume the rest of the propane will combust. Only 8-12% of the propane is taken up by air (2 parts air at 21% O2 content) (3.5 O2 + C3H8 --> 3 CO + 4 H2O) (5 O2 + C3H8 --> 3 CO2 + 4 H2O). The other ~90% undergoes about 7x gas molecule expansion, netting about another 2x from propane combustion. There's also going to be some decomposition of the AN, yielding even more gas. I would have to guess that there may well be some positive feedback runaway here, as decomposition drives higher P/T. At this point my ability to estimate quickly runs out. Better estimates would require more work.

There's one part of the story that could change much of the analysis, which is that there's liquid propane in the charge. If that's the case, this whole thing would act far more like AN/FO than just AN.
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[*] posted on 26-6-2011 at 18:03


The real problem is that the specific heat of gases is very low compared
to that of solids. The mass of the AN will absorb most of the heat, the flame may not even propagate in between the prills (especially if air is used). There is probably
only enough heat generated from the propane burning with the air to heat the
mass of AN up a few degrees. Think of a car engine, it must be running for several
minutes (the cylinders fire thousands of times) before it warms up.
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[*] posted on 26-6-2011 at 19:46


Quote:
There is probably
only enough heat generated from the propane burning with the air to heat the
mass of AN up a few degrees.


'there is only enough energy in this match to warm this sheet of newsprint by a few degrees... I can't start a fire...'

Back of the envelope says that yes, there's only about 30J per g of AN (from the combustion of the propane), so if the heat were transferred instantly and evenly to *all* of the AN, it would just get warm. But the prills have an outside surface, the combustion is rapid, and once the decomposition of the AN sets in more energy gets released. And what does the *specific heat* of gases have to do with anything here, it's the heat of combustion that's important.
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[*] posted on 27-6-2011 at 08:26


Quote: Originally posted by gregxy  
The real problem is that the specific heat of gases is very low compared
to that of solids. The mass of the AN will absorb most of the heat, the flame may not even propagate in between the prills (especially if air is used).
If this were a steady-state heat problem, I'd agree, but it's not. The nominal time frame we have here is only about a 1 ms. There's not enough time to for significant heat flow into the prills. In addition, we've got to take into account surface melting dynamics here. There's some energy that goes into the phase change, true, but far more important is that the solid-liquid interface is going to be at the melting point of AN. Heat flow is proportional to temperature difference, and the melting point of AN is only 170 &deg;C, which is way closer to ambient temperature than the flame temperature. So the bulk of the combustion heat is going to be partitioned, in the nominal 1 ms time frame, vastly in favor of the surface. The decomposition temperature is only a bit more, 210 &deg;C, so the partition between liquid and decomposition further favors decomposition.

If we were in a steady-state situation, then, yes, we would compare heat capacities and total masses, or alternately specific heats, densities, and fill ratios. The heat energy partition, though, would be very different in such a situation. It's definitely easier, though, to calculate the heat energy partition in the steady state. Estimating the surface dynamics is still not hugely difficult. The simplest model to consider first is one-dimensional along an axis transverse to the surface. Consider the left side as a heat bath at ambient temperature (300 K) and the right one as a one at flame temperature (900 K or so). Model the solid as everything less than zero. Model the liquid as the interval (0, a). Model recently decomposed gas as leidenfrost-effect layer in the interval (a, a+b). We have to just guess how thick that layer is the value of b, but a micron seems about right, typical of boundary layers. Boundary conditions are that T<sub>0</sub> = melting point, T<sub>a</sub> = decomposition point, T<sub>b</sub> = flame temperature. There's a full differential equation to write down, but to estimate it we look for a steady-state solution for this surface process. What we want to calculate is the thickness of the liquid layer (a) and the velocities of the boundaries at 0 and a. In the present steady-state problem, we assume that these velocities are identical. We know that the form of the solution <0 is a decaying exponential and that in the intervals is a line. At the boundary points at 0 and a, there's a conservation of energy equation in rate form, where the difference in energy in and energy out (the one-sided first derivatives at each side of the boundary point), the excess energy goes into phase change, and that yields a velocity for the boundary point. We have two boundary equations, two unknowns (a velocity and a thickness), so we're properly constrained.

I'm not going to work out the details myself, but I can check calculations of someone who wants to.
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[*] posted on 27-6-2011 at 09:00


Quote: Originally posted by dann2  


The original question has only been answered (probably speculation) my one person in the entire thread. (Thank you freedompyro). Dann2 said he haddent a clue.

We have had a discussion regarding:
Difficulty of removing valves from Propane tanks.



Well Boooo-Hooo. Sorry the rest of us are neither magicians
or Gods and thus will have to remove la valve. As previously
noted by Id — on 32# and larger tanks this is next to impossible.

And the question me thinks yet to be answered ...
does/why would propane react with ammonium nitrate?
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[*] posted on 27-6-2011 at 10:11


I worked through the math for another problem, (solved the one dimensional time dependent heat flow equation with two domains, one gas and one solid). What I found was that heat conduction in the solid and its high specific heat as compared to the gas prevent even the surface from getting very hot until you have heated the bulk.


Furthermore even sensitive explosives do not respond to this kind of stimula. If you pass ETN crystals through a flame, they don't explode, the sharp corners just melt a little. Granted the pressure is much lower with flame, but you can keep them in the flame for several seconds. No decomposition occurs until the entire mass is melted.

Finally detonation is a strong compression wave traveling in a solid.
Acoustic waves in a gas don't transfer that much energy to a solid, instead the waves bounce off. Detonation between blocks of HEs will not pass through more than a few cm of air.

[Edited on 27-6-2011 by gregxy]
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[*] posted on 27-6-2011 at 11:27


Quote: Originally posted by gregxy  
I worked through the math for another problem, (solved the one dimensional time dependent heat flow equation with two domains, one gas and one solid). What I found was that heat conduction in the solid and its high specific heat as compared to the gas prevent even the surface from getting very hot until you have heated the bulk.
These behaviors depend upon both heat capacity and the heat conductivity (roughly, the R-value). Also, in this situation, we have a liquid phase layer with much better heat conductivity than the solid and rather different heat capacity. Therefore, I'm not at all convinced that the qualitative aspects of a two-domain model with no phase changes is simple analogue of a four-domain model with two phase changes.

Disclaimer: The model I presented used a linear solution in the liquid and gas. This is equivalent to using an infinite heat conductivity value. As such, it represents a kind of limit case. In practice, finite values for this will partition even more heat into the gas-boundary and liquid-surface phases. It will tend, if I'm thinking correctly, to heat the liquid layer more and increase the decomposition rate.

The whole point of doing the four-domain model is to estimate the rate of AN decomposition. If it's at all significant, it'll mean that the "passing ETN through flame" example is insufficiently alike to give analogous results. It's a difference between pressure of 1 atm vs. perhaps 200 atm or more, so it can hardly be neglected.
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[*] posted on 1-7-2011 at 23:12


Just so you guys know... In the detonation velocities document there is a reference that says Propane/Oxygen has a velocity of 1871m/s... The source for this information is D362? Edit: Source is PATR 2700 D362

Doesn't indicate what density... Can't be liquid propane/oxygen... Would be much more velocity than that.

[Edited on 2-7-2011 by freedompyro]
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